# Volume change in positive deviation from Raoult's Law

Let us take a mixture of ethanol and acetone as an example. When ethanol is mixed with acetone, the hydrogen bonding between ethanol molecules gets disturbed as actone molecules get in between the ethanol molecules. This results in $$\Delta_\text{mix} H\gt0$$ as heat must be supplied to make the solution from its constituents (as attraction is less) but I am not sure about $$\Delta_\text{mix} V$$. This page says it is greater than zero because volume expands on mixing. But I think it should be less than zero because the observed vapour pressure is more than that predicted on the basis of theoretical calculations, so a greater amount of solution will be present in the form of vapour, hence volume of solution will be less (than the sum of the volumes of its constituents). Where am I wrong?

• An intuitive explanation can be found on the page you referenced. The lessened attraction between the constituents effectively means that the average "intermolecular distance" would be greater after mixing than it was before, hence the increased volume. The vapor pressures before and after mixing are not directly relevant here. – voffch Mar 8 at 13:59
• I agree with the increased intermolecular distance, but more vapour pressure means more amount of vapour over solution. As total amount of solution + vapour is constant, this means less solution, hence less volume of solution, so there are two conflicting processes which one dominates? – drake01 Mar 8 at 14:20
• @drake01 By the volume change is meant the total volume after mixing, at the same temperature versus sum of volumes of the same amount of liquids before mixing. It should be obvious that evaporation changes the volume, as the there is less of liquids. – Poutnik Sep 14 at 7:32

but I am not sure about $$\Delta_{mix}V$$. [...] I think it should be less than zero because the observed vapour pressure is more than that predicted on the basis of theoretical calculations
$$\Delta_{mix}V$$ refers to the change in the volume of the liquid with mixing, compared to the pure components. The vapour plays no role in this computation. Note also that the lower vapour pressure relative to the pure components means that the free energy of the liquid mixture is lower than the arithmetic average over the free energies of the pure components. This can certainly be interpreted as being due to stronger association between the components in the mixture than in the average of the pure liquids, but does not necessarily imply a specific direction of change (positive or negative) in volume.